Answer :
To determine the least number [tex]\( n \)[/tex] such that the [tex]\( n \)[/tex]th term of the geometric sequence is greater than the corresponding term in the arithmetic sequence, we first need to define both sequences clearly.
Arithmetic Sequence:
- First term [tex]\( a \)[/tex] is 600.
- Common difference [tex]\( d \)[/tex] is 400.
- General formula for the [tex]\( n \)[/tex]th term: [tex]\( a_n = a + (n-1)d \)[/tex].
Geometric Sequence:
- First term [tex]\( g \)[/tex] is 2.
- Common ratio [tex]\( r \)[/tex] is 2.
- General formula for the [tex]\( n \)[/tex]th term: [tex]\( g_n = g \times r^{n-1} \)[/tex].
We need to find the smallest [tex]\( n \)[/tex] such that [tex]\( g_n > a_n \)[/tex], i.e.,
[tex]\[ 2 \times 2^{n-1} > 600 + (n-1) \times 400. \][/tex]
Let's substitute the specific values for each [tex]\( n \)[/tex] to find the point where the geometric sequence term exceeds the arithmetic sequence term:
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{0} = 2, \][/tex]
[tex]\[ a_n = 600 + (1-1) \times 400 = 600. \][/tex]
Clearly, [tex]\( 2 < 600 \)[/tex].
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{1} = 4, \][/tex]
[tex]\[ a_n = 600 + (2-1) \times 400 = 1000. \][/tex]
Clearly, [tex]\( 4 < 1000 \)[/tex].
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{2} = 8, \][/tex]
[tex]\[ a_n = 600 + (3-1) \times 400 = 1400. \][/tex]
Clearly, [tex]\( 8 < 1400 \)[/tex].
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{3} = 16, \][/tex]
[tex]\[ a_n = 600 + (4-1) \times 400 = 1800. \][/tex]
Clearly, [tex]\( 16 < 1800 \)[/tex].
5. For [tex]\( n = 5 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{4} = 32, \][/tex]
[tex]\[ a_n = 600 + (5-1) \times 400 = 2200. \][/tex]
Clearly, [tex]\( 32 < 2200 \)[/tex].
6. For [tex]\( n = 6 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{5} = 64, \][/tex]
[tex]\[ a_n = 600 + (6-1) \times 400 = 2600. \][/tex]
Clearly, [tex]\( 64 < 2600 \)[/tex].
7. For [tex]\( n = 7 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{6} = 128, \][/tex]
[tex]\[ a_n = 600 + (7-1) \times 400 = 3000. \][/tex]
Clearly, [tex]\( 128 < 3000 \)[/tex].
8. For [tex]\( n = 8 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{7} = 256, \][/tex]
[tex]\[ a_n = 600 + (8-1) \times 400 = 3400. \][/tex]
Clearly, [tex]\( 256 < 3400 \)[/tex].
9. For [tex]\( n = 9 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{8} = 512, \][/tex]
[tex]\[ a_n = 600 + (9-1) \times 400 = 3800. \][/tex]
Clearly, [tex]\( 512 < 3800 \)[/tex].
10. For [tex]\( n = 10 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{9} = 1024, \][/tex]
[tex]\[ a_n = 600 + (10-1) \times 400 = 4200. \][/tex]
Clearly, [tex]\( 1024 < 4200 \)[/tex].
11. For [tex]\( n = 11 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{10} = 2048, \][/tex]
[tex]\[ a_n = 600 + (11-1) \times 400 = 4600. \][/tex]
Clearly, [tex]\( 2048 < 4600 \)[/tex].
12. For [tex]\( n = 12 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{11} = 4096, \][/tex]
[tex]\[ a_n = 600 + (12-1) \times 400 = 5000. \][/tex]
Clearly, [tex]\( 4096 < 5000 \)[/tex].
13. For [tex]\( n = 13 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{12} = 8192, \][/tex]
[tex]\[ a_n = 600 + (13-1) \times 400 = 5400. \][/tex]
Now, [tex]\( 8192 > 5400 \)[/tex].
Therefore, the least number [tex]\( n \)[/tex] such that the [tex]\( n \)[/tex]th term of the geometric sequence is greater than the corresponding term in the arithmetic sequence is [tex]\( n = 13 \)[/tex].
Arithmetic Sequence:
- First term [tex]\( a \)[/tex] is 600.
- Common difference [tex]\( d \)[/tex] is 400.
- General formula for the [tex]\( n \)[/tex]th term: [tex]\( a_n = a + (n-1)d \)[/tex].
Geometric Sequence:
- First term [tex]\( g \)[/tex] is 2.
- Common ratio [tex]\( r \)[/tex] is 2.
- General formula for the [tex]\( n \)[/tex]th term: [tex]\( g_n = g \times r^{n-1} \)[/tex].
We need to find the smallest [tex]\( n \)[/tex] such that [tex]\( g_n > a_n \)[/tex], i.e.,
[tex]\[ 2 \times 2^{n-1} > 600 + (n-1) \times 400. \][/tex]
Let's substitute the specific values for each [tex]\( n \)[/tex] to find the point where the geometric sequence term exceeds the arithmetic sequence term:
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{0} = 2, \][/tex]
[tex]\[ a_n = 600 + (1-1) \times 400 = 600. \][/tex]
Clearly, [tex]\( 2 < 600 \)[/tex].
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{1} = 4, \][/tex]
[tex]\[ a_n = 600 + (2-1) \times 400 = 1000. \][/tex]
Clearly, [tex]\( 4 < 1000 \)[/tex].
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{2} = 8, \][/tex]
[tex]\[ a_n = 600 + (3-1) \times 400 = 1400. \][/tex]
Clearly, [tex]\( 8 < 1400 \)[/tex].
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{3} = 16, \][/tex]
[tex]\[ a_n = 600 + (4-1) \times 400 = 1800. \][/tex]
Clearly, [tex]\( 16 < 1800 \)[/tex].
5. For [tex]\( n = 5 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{4} = 32, \][/tex]
[tex]\[ a_n = 600 + (5-1) \times 400 = 2200. \][/tex]
Clearly, [tex]\( 32 < 2200 \)[/tex].
6. For [tex]\( n = 6 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{5} = 64, \][/tex]
[tex]\[ a_n = 600 + (6-1) \times 400 = 2600. \][/tex]
Clearly, [tex]\( 64 < 2600 \)[/tex].
7. For [tex]\( n = 7 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{6} = 128, \][/tex]
[tex]\[ a_n = 600 + (7-1) \times 400 = 3000. \][/tex]
Clearly, [tex]\( 128 < 3000 \)[/tex].
8. For [tex]\( n = 8 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{7} = 256, \][/tex]
[tex]\[ a_n = 600 + (8-1) \times 400 = 3400. \][/tex]
Clearly, [tex]\( 256 < 3400 \)[/tex].
9. For [tex]\( n = 9 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{8} = 512, \][/tex]
[tex]\[ a_n = 600 + (9-1) \times 400 = 3800. \][/tex]
Clearly, [tex]\( 512 < 3800 \)[/tex].
10. For [tex]\( n = 10 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{9} = 1024, \][/tex]
[tex]\[ a_n = 600 + (10-1) \times 400 = 4200. \][/tex]
Clearly, [tex]\( 1024 < 4200 \)[/tex].
11. For [tex]\( n = 11 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{10} = 2048, \][/tex]
[tex]\[ a_n = 600 + (11-1) \times 400 = 4600. \][/tex]
Clearly, [tex]\( 2048 < 4600 \)[/tex].
12. For [tex]\( n = 12 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{11} = 4096, \][/tex]
[tex]\[ a_n = 600 + (12-1) \times 400 = 5000. \][/tex]
Clearly, [tex]\( 4096 < 5000 \)[/tex].
13. For [tex]\( n = 13 \)[/tex]:
[tex]\[ g_n = 2 \times 2^{12} = 8192, \][/tex]
[tex]\[ a_n = 600 + (13-1) \times 400 = 5400. \][/tex]
Now, [tex]\( 8192 > 5400 \)[/tex].
Therefore, the least number [tex]\( n \)[/tex] such that the [tex]\( n \)[/tex]th term of the geometric sequence is greater than the corresponding term in the arithmetic sequence is [tex]\( n = 13 \)[/tex].
